Question

$(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})...(1+\frac{(2x+1)}{n^2})={(n+1)}^2.$

Answer 1

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left(1+\frac{\left(2n+1\right)}{n^2}\right)$

Here $a_n=1+\frac{2n+1}{n^2}$

$=1+\frac{2}{n}+\frac{1}{n^2}$

$={\left(1+\frac{1}{n}\right)}^2$

$={\left(\frac{n+1}{n}\right)}^2$

Therefore, $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left(1+\frac{\left(2n+1\right)}{n^2}\right)$$=\prod _1^n{a}_n$

$=\prod _1^n{\left(\frac{n+1}{n}\right)}^2$

$=\prod _1^n\frac{{\left(n+1\right)}^2}{n^2}$

$=\frac{2^2}{1}\times \frac{3^2}{2^2}\times \frac{4^2}{3^2}\times \frac{5^2}{4^2}...\frac{{\left(n+1\right)}^2}{n^2}$

$={\left(n+1\right)}^2$